Optimal Grid Path Protocol — Problem Statement & Solution Guide
Problem Description
Your task is to calculate the optimal grid path using the Subsequence Verification methodology. Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Grid Path Protocol"
WHY DOES IT MATTER?
Subsequence verification appears in autocomplete engines, DNA sequence analysis, and permission‑checking pipelines where order matters but contiguity does not. Mastering the two‑pointer pattern equips engineers to solve a wide class of linear‑order problems efficiently.
OPTIMIZATION CHALLENGE
The key insight is that you never need to reconsider a character once it has been skipped; the earliest possible match for each target character yields a globally optimal solution, eliminating exponential backtracking.
REAL-WORLD CONNECTION
Think of a train moving along a track (the source string) while a conductor checks off stations (the target string) in order. The conductor only steps forward when the next required station appears, mirroring the greedy pointer advance in the algorithm.
During an interview, write the two‑pointer loop first, then immediately discuss edge cases (empty strings, repeated characters) and, if time permits, mention the preprocessing‑binary‑search extension for massive query sets.
COMPLEXITY AT A GLANCE
O(|s| + |t|)O(1)Core Theory — Why This Approach?
The optimal solution hinges on the two‑pointer (or greedy) subsequence verification technique. By iterating through the source string once and advancing a pointer in the target string only when characters match, we can decide in linear time whether the target is a subsequence of the source. This works because the relative order of characters is the only constraint; any matching character can be safely taken without affecting future possibilities. Naïve approaches—such as checking every possible combination of indices or using nested loops—exhibit O(|source|·|target|) time, which quickly becomes infeasible when both strings approach 10^5 or more characters. The optimal paradigm reduces the problem to a single pass, guaranteeing O(|source|+|target|) time and O(1) auxiliary space, satisfying the stringent constraints of modern coding interviews and large‑scale systems.
When the problem extends to multiple queries (e.g., many target strings against a single large source), preprocessing the source into a positional index (an array of vectors for each alphabet character) enables binary‑search‑based jumps, preserving logarithmic per‑query cost. This extension illustrates the broader principle: transform a linear scan into a searchable structure to handle massive query volumes while keeping overall complexity under control.
Interview Questions on This Problem
Q1How would you verify if string t is a subsequence of string s in O(|s|+|t|) time, and why does this guarantee optimality?
Use two pointers i and j starting at 0 for s and t. Increment i each step; when s[i] == t[j], increment j. If j reaches |t|, t is a subsequence. This runs in a single pass over s, giving O(|s|+|t|) which is optimal because any algorithm must inspect each character of s at least once.
Q2A fintech platform needs to process 10^5 subsequence queries against a static transaction log of length 10^6. Which data structure would you employ and what is the per‑query complexity?
Preprocess the log into a 26‑element array of vectors storing indices of each character. For each query, walk through its characters and binary‑search the next index greater than the previous match. This yields O(|query|·log N) per query, where N is the log length.
Q3In a high‑growth startup, you must extend the subsequence check to support wildcard characters that can match any single letter. How does this affect the algorithmic approach?
Treat a wildcard as an automatic match without advancing the source pointer; simply move to the next character in the pattern. The two‑pointer scan still works, preserving O(|s|+|t|) time, because wildcards never force backtracking.
Examples
Input
[7, 12, 17, 7]
Output
43
Explanation: Step-by-step: The input array is [7, 12, 17, 7]. To calculate the optimal grid path, we need to find the sum of all elements in the array. The sum is 7 + 12 + 17 + 7 = 43.
Input
[2, 4]
Output
6
Explanation: Step-by-step: The input array is [2, 4]. To calculate the optimal grid path, we need to find the sum of all elements in the array. The sum is 2 + 4 = 6.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Use a two‑pointer scan (or binary‑search on a pre‑processed index) to achieve linear or logarithmic per‑query time.
Brute Force Approach
Try every possible combination of indices in the source string to match the target, leading to exponential or O(|s|·|t|) time.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
if not nums:
return 0
return sum(nums)function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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